How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
240o
210o
150o
60o
Correct answer is B
Using sinØ = \(\frac{6√3}{12}\) → \(\frac{√3}{2}\)
sinØ = \(\frac{√3}{2}\) or 60°
The bearing of Z from X = [270 - 60]° → 210°
If in the diagram, FG is parallel to KM, find the value of x
75o
95o
105o
125o
Correct answer is B
No explanation has been provided for this answer.
\(\frac{800\pi}{9}\)km
\(\frac{800\sqrt{3\pi}}{9}\)km
800\(\pi\) km
800\(\sqrt{3\pi}\) km
Correct answer is C
Angular difference (\(\theta\))= 25° + 20° = 45°
\(\alpha\) = common latitude = 60°
\(S = \frac{\theta}{360°} \times 2\pi R \cos \alpha\)
\(S = \frac{45°}{360°} \times 2 \pi \times 6400 \times \cos 60°\)
= \(\frac{6400\pi}{8} = 800\pi km\)
If the angles of quadrilateral are (p + 10)°, (2p - 30)°, (3p + 20)° and 4p°, find p.
63
40
36
28
Correct answer is C
The sum of angles in a quadrilateral = 360°
\(\therefore (p + 10) + (2p - 30) + (3p + 20) + 4p = 360\)
\(10p = 360° \implies p = \frac{360}{10} = 36°\)
PPT
pp-1
qp
pp
Correct answer is A
p = \(\begin{vmatrix} 0 & 3 & 0 \\ 2 & 1 & 3\\ 4 & 2 & 2 \end{vmatrix}\)
Q = \(\begin{vmatrix} 0 & 2 & 4 \\ 3 & 1 & 2\\ 0 & 3 & 2 \end{vmatrix}\) = pT
pq = ppT